Objective-C中的二叉树

本文介绍了Objective-C中的二叉树的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在学习算法和数据结构，并且训练我正在尝试使用objective-c设计和实现二叉树。

I am learning algorithms and data structures and to train I am trying to design and implement a binary tree using objective-c.

到目前为止，我有以下课程：

So far I have the following Classes:

main - 用于测试

节点 - 树节点

BinaryTree - 适用于与树相关的所有方法

main - for testing
Node - node of tree
BinaryTree - for all methods related to the tree

其中一个我实现的 BinaryTree 类中的第一个方法是 insertNode：forRoot：。

One of the first methods in BinaryTree class I implemented is insertNode:forRoot:.

- (void)insertNodeByRef:(Node **)node forRoot:(Node **)root{

    if (head == NULL) {
        head = *node;
    }
    // Case 2 root is null so can assign the value of the node to it
    if (root == NULL) {
        root = node;
    } else {
        if (node.data > root.data) { // to the right
            [self insertNode:node forRoot:root.right];
        } else if (node.data < root.data) { //or to the left
            [self insertNode:node forRoot:root.left];
        }
    }
}

<$> c $ c>节点类看起来像：

@interface Node : NSObject

@property(nonatomic, assign) int data;
@property(nonatomic, strong) Node * right;
@property(nonatomic, strong) Node * left;

@end

我的问题是我不知道怎么做如果我将Node作为引用传递，则访问Node类成员变量。每当我尝试访问节点属性（如数据，左或右）时，我收到以下错误消息：

My problem is that I don't know how to access the Node class member variables if I am passing Node as a reference. Whenever I try to access the node properties (like data, left or right) I am getting the following error message:

Member reference base type 'Node *__autoreleasing *' is not a structure or union

所以我的问题是：
如何访问这些属性（数据，左或右）并使用它们存储int数据或引用另一个节点？

So my questions is:how can I access those properties (data, left or right) and use them to store either int data or reference to another Node?

希望它有意义。谢谢！

推荐答案

您的代码混合了两种常见的任务方法，因此存在问题。您还使用抽象数据类型（ADT）类型方法，而不是面向对象的方法，因此有三种方法需要考虑。

Your code is mixing two common approaches to the task, hence the problem. You are also using an abstract data type (ADT) type approach, rather than an object-oriented one, so there are three approaches to consider.

在两种ADT方法中，您的树由对其根的引用表示，在Objective-C中，这可能存储在实例变量中：

In both ADT approaches your tree is represented by a reference to its root, in Objective-C this is probably stored in an instance variable:

Node *TreeRoot;

另请注意，这两种算法都使用字段引用， a-> b ，而不是属性引用， ab - 这是因为前者引用变量而第二种算法需要传递对变量的引用。

Note also that both of these algorithms use field references, a->b, rather than property references, a.b - this is because the former references a variable and the second algorithm requires passing a reference to a variable.

功能性ADT：按值传递并分配结果

在此方法中，将节点插入树中，并返回已修改的树，该树将被分配回来，例如插入节点 nodeToInsert 的顶级调用将是：

In this approach a node is inserted into a tree and a modified tree is returned which is assigned back, e.g. the top-level call to insert a Node nodeToInsert would be:

TreeRoot = insertNode(nodeToInsert, TreeRoot);

和 insertNode 函数如下所示： / p>

and the insertNode function looks like:

Node *insertNode(Node *node, Node *root)
{
   if(root == nil)
   {  // empty tree - return the insert node
      return node;
   }
   else
   {  // non-empty tree, insert into left or right subtree
      if(node->data > root->data) // to the right
      {
        root->right = insertNode(node, root->right);
      }
      else if(node->data < root->data)//or to the left
      {
        root->left = insertNode(node, root->left);
      }
      // tree modified if needed, return the root
      return root;
   }
}

请注意，在这种方法中，如果是非-empty（sub）tree算法对变量执行冗余赋值 - 赋值是变量中已有的值...因此有些人更喜欢：

Note that in this approach in the case of a non-empty (sub)tree the algorithm performs a redundant assignment into a variable - the assigned value is what is already in the variable... Because of this some people prefer:

程序ADT：传递参考

在这种方法中，保存（子）树根的变量是通过 - 引用，而不是传递的值，并根据需要由被调用的过程进行修改。例如。顶级调用将是：

In this approach the variable holding the root of the (sub)tree is passed-by-reference, rather than its value being passed, and is modified by the called procedure as needed. E.g. the top-level call would be:

insertNode(nodeToInsert, &TreeRoot); // & -> pass the variable, not its value

和 insertNode 过程如下所示：

void insertNode(Node *node, Node **root)
{
   if(*root == nil)
   {  // empty tree - insert node
      *root = node;
   }
   else
   {  // non-empty tree, insert into left or right subtree
      Node *rootNode = *root;
      if(node->data > rootNode->data) // to the right
      {
        insertNode(node, &rootNode->right);
      }
      else if(node->data < rootNode->data)//or to the left
      {
        insertNode(node, &root->left);
      }
   }
}

您现在可以看到您的方法是上述两种方法的混合物。两者都是有效的，但是当您使用Objective-C时，采用第三种方法可能更好：

You can now see that your method is a mixture of the above two approaches. Both are valid, but as you are using Objective-C it might be better to take the third approach:

面向对象的ADT

这是程序ADT的变体 - 而不是将变量传递给过程变量，现在称为对象，拥有方法更新自己。这样做意味着您必须在调用插入节点之前测试空（子）树，而前两个方法在调用中进行测试。所以现在我们在 Node 中有方法：

This is a variation of the procedural ADT - rather than pass a variable to a procedure the variable, now called an object, owns a method which updates itself. Doing it this way means you must test for an empty (sub)tree before you make a call to insert a node, while the previous two approaches test in the call. So now we have the method in Node:

- (void) insert:(Node *)node
{
   if(node.data > self.data) // using properties, could also use fields ->
   {
      if(self.right != nil)
         [self.right insert:node];
      else
         self.right = node;
   }
   else if(node.data < rootNode.data)
   {
      if(self.left != nil)
         [self.left insert:node];
      else
         self.left = node;
   }
}

你还需要更改顶级电话来做空树的相同测试：

You also need to change the top level call to do the same test for an empty tree:

if(TreeRoot != nil)
   [TreeRoot insert:nodeToInsert];
else
   TreeRoot = nodeToInsert;

最后一点 - 如果您使用MRC而不是ARC或GC进行内存管理我需要插入适当的保留/释放电话。

And a final note - if you are using MRC, rather than ARC or GC, for memory management you'll need to insert the appropriate retain/release calls.

希望能帮助你解决问题。

Hope that helps you sort things out.

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